Optimal. Leaf size=1392 \[ -\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )^2 (-b)^{5/2}}{2 f^2}-\frac {(c+d x) \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) (-b)^{5/2}}{f}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right ) (-b)^{5/2}}{f^2}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2 \left (\sqrt {b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right ) (-b)^{5/2}}{2 f^2}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (-\frac {2 \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right ) (-b)^{5/2}}{2 f^2}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2}{\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1}\right ) (-b)^{5/2}}{f^2}+\frac {d \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right ) (-b)^{5/2}}{2 f^2}-\frac {d \text {Li}_2\left (1-\frac {2 \left (\sqrt {b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right ) (-b)^{5/2}}{4 f^2}-\frac {d \text {Li}_2\left (\frac {2 \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}+1\right ) (-b)^{5/2}}{4 f^2}+\frac {d \text {Li}_2\left (1-\frac {2}{\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1}\right ) (-b)^{5/2}}{2 f^2}+\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )^2}{2 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+\frac {2 b^{5/2} d \tan ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{3 f^2}+\frac {b^{5/2} (c+d x) \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{f}+\frac {2 b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{3 f^2}-\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b}}{\sqrt {b}-\sqrt {b \tanh (e+f x)}}\right )}{f^2}+\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b}}{\sqrt {b}+\sqrt {b \tanh (e+f x)}}\right )}{f^2}-\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b} \left (\sqrt {-b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{2 f^2}-\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b} \left (\sqrt {-b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{2 f^2}-\frac {b^{5/2} d \text {Li}_2\left (1-\frac {2 \sqrt {b}}{\sqrt {b}-\sqrt {b \tanh (e+f x)}}\right )}{2 f^2}-\frac {b^{5/2} d \text {Li}_2\left (1-\frac {2 \sqrt {b}}{\sqrt {b}+\sqrt {b \tanh (e+f x)}}\right )}{2 f^2}+\frac {b^{5/2} d \text {Li}_2\left (1-\frac {2 \sqrt {b} \left (\sqrt {-b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{4 f^2}+\frac {b^{5/2} d \text {Li}_2\left (1-\frac {2 \sqrt {b} \left (\sqrt {-b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{4 f^2}-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2} \]
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Rubi [F] time = 0.13, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin {align*} \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx &=-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx+\frac {(2 b d) \int (b \tanh (e+f x))^{3/2} \, dx}{3 f}\\ &=-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx+\frac {\left (2 b^3 d\right ) \int \frac {1}{\sqrt {b \tanh (e+f x)}} \, dx}{3 f}\\ &=-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx-\frac {\left (2 b^4 d\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-b^2+x^2\right )} \, dx,x,b \tanh (e+f x)\right )}{3 f^2}\\ &=-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx-\frac {\left (4 b^4 d\right ) \operatorname {Subst}\left (\int \frac {1}{-b^2+x^4} \, dx,x,\sqrt {b \tanh (e+f x)}\right )}{3 f^2}\\ &=-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx+\frac {\left (2 b^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \tanh (e+f x)}\right )}{3 f^2}+\frac {\left (2 b^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \tanh (e+f x)}\right )}{3 f^2}\\ &=\frac {2 b^{5/2} d \tan ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{3 f^2}+\frac {2 b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{3 f^2}-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx\\ \end {align*}
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Mathematica [F] time = 39.85, size = 0, normalized size = 0.00 \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx \]
Verification is Not applicable to the result.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \left (b \tanh \left (f x + e\right )\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.22, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right ) \left (b \tanh \left (f x +e \right )\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \left (b \tanh \left (f x + e\right )\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,\mathrm {tanh}\left (e+f\,x\right )\right )}^{5/2}\,\left (c+d\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tanh {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (c + d x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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